On the buoyancy of ships and stones entering the sea from the river I know that because the density of river water is different from that of sea water, the water level of ship entering the sea rises But also into the sea, the stone has always been at the bottom. Has its density changed?

Buoyancy = density of water * g * volume of drainage
The density of the ship entering the sea water increases, the drainage volume decreases, and the buoyancy is equal to gravity
How the density of the stone into the sea is large, the drainage volume remains unchanged, the buoyancy becomes larger, but it is still small and gravity

The stones in the river are washed into the sea. Will the buoyancy change

Buoyancy = density * g * volume of discharged water seawater density is large. You should know that the buoyancy of ships sailing into the sea from rivers does not change = gravity, but the volume of discharged water is small, that is, it floats high, so it changes. The volume of discharged water is the same, and the seawater density is large, and the buoyancy is large
It will change. First look at the f float= ρ Liquid V discharge g, that is, the buoyancy is equal to the density of the liquid multiplied by the volume of the discharged liquid multiplied by 10. The stone must be completely submerged in the water in the river and in the sea water. Therefore, the density of the discharged liquid is equal to the volume of the stone. That volume is fixed, and the density of the sea water is greater than that of the river. Therefore, the buoyancy becomes larger
In popular terms, people can float in the dead sea because the dead sea has a large salt content and density, so it has a large buoyancy. It will change. First, let's look at f floating= ρ Liquid V discharge g, that is, the buoyancy is equal to the density of the liquid multiplied by the volume of the discharged liquid multiplied by 10. The stone must be completely submerged in the water in the river and in the sea water. Therefore, the density of the discharged liquid is equal to the volume of the stone. That volume is fixed, and the density of the sea water is greater than that of the river. Therefore, the buoyancy becomes larger
In popular terms, people can float in the dead sea because the dead sea has high salt content and density, so it has high buoyancy

Press the function y = sin2x as a vector a=（-π 6, 1) the analytical formula of the function after translation is __

Press the function y = sin2x as a vector
a=（-π
6,1) after translation, the analytical formula of the function y = sin2 (x + π) is obtained
6) + 1, i.e. y = sin (2x + π)
3)+1，
So the answer is: y = sin (2x + π
3)+1．

After translating the image of function y = sin2x according to vector a = (- π / 4,1), the analytical formula of the image obtained is the square of y = 2cosx. How to translate?

y =sin[2(x+π/4)]+1
=sin(2x+2/π)+1
=cos2x+1
=cosx ²- sinx ²+ one
=cosx ²- sinx ²+ cosx ²+ sinx ²
=2cosx ²

Translate the image of function y = sin2x-1 along the vector a = (π \ 4,1), then the analytical formula of the function corresponding to the translated image is

y+1=sin[2(x+π/4)]-1
have to
y=sin(2x+π/2)-2=cos2x-2

Let the function f (x) = cosx SiNx. After translating the function image of F (x) according to the vector a = (m, 0) (M > 0), we just get the image of the function y = - f '(x) Then the value of M can be?

F (x) = cosx SiNx = √ 2 (√ 2 / 2cosx - √ 2 / 2sinx) = √ 2 (COS π / 4cosx sin π / 4sinx) = √ 2cos (x + π / 4) f '(x) = - √ 2Sin (x + π / 4) y = - F' (x) = √ 2Sin (x + π / 4) √ 2cos (x + π / 4) → translate (π / 2 + 2K π) units to the right to get →√ 2Sin (x + π / 4) (M > 0), so m =

On the buoyancy of ships and stones entering the sea from the river I know that because the density of river water is different from that of sea water, the water level of ship entering the sea rises But also into the sea, the stone has always been at the bottom. Has its density changed?